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Make three non-empty sets with negative, positive and 0 products

  • Difficulty Level : Medium
  • Last Updated : 22 Jun, 2021

You are given an array of n distinct integers. Your task is to divide this array into three non-empty sets so as the following conditions hold: 
 

  1. The first set should contain the elements such that their product should be less than 0.
  2. The Second set should contain the elements such that their product should be greater than 0.
  3. The Third set should contain the elements such that their product should be equal to 0.

Notes:- 
 

  1. In the give array, each number must occur in exactly one set.
  2. It may be assumed that we can always make three sets (there is at-least one negative element and one 0 in input array).

Examples: 
 

Input : 4
        arr[] = -1 -2 -3 0
Output :  -1
          -3 -2 
           0
In this example, product of first
set is negative, product of second
set is positive and product of third
set is 0.

 

In this problem we just need to Split input data into 3 vectors: first will contain negative numbers, second positive numbers, third zeroes. If size of first vector is even move one number from it to the third vector. If second vector contains only 1, then move two numbers from first vector to the second vector.
 



C++




// CPP program to make three non-empty sets
// as per the given conditions.
#include <bits/stdc++.h>
using namespace std;
 
void makeSets(int arr[], int n)
{
    vector<int> first, second, third;
 
    // insert number equal to 0 to third set.
    // numbers greater than 0 to second set.
    // insert numbers less than 0 to first set.
    for (int i = 0; i < n; i++) {
        if (arr[i] == 0)
            third.push_back(arr[i]);
        if (arr[i] > 0)
            second.push_back(arr[i]);
        if (arr[i] < 0)
            first.push_back(arr[i]);
    }
 
    if (first.size() == 0 || third.size() == 0)
    {
        cout << "Not Possible";
        return;
    }
 
    // if second set is empty.
    if (second.size() == 0) {
        for (int i = 0; i < 2; i++)
        {
            second.push_back(first.back());
            first.pop_back();
        }
    }
 
    // if length of first set is even.
    if (first.size() % 2 == 0) {
        third.push_back(first.back());
        first.pop_back();
    }
 
    // output the first set elements.
    for (int i = 0; i < first.size(); i++)
        cout << first[i] << " ";   
     
    // output the second set elements.
    cout << endl;
    for (int i = 0; i < second.size(); i++)
        cout << second[i] << " ";   
 
    // output the third set elements.
    cout << endl;
    for (int i = 0; i < third.size(); i++)
        cout << third[i] << " ";   
}
 
// Driver Function
int main()
{
    int arr[] = { -1, 2, 0 };
    int n = sizeof(arr) / sizeof(arr[0]);
    makeSets(arr, n);
    return 0;
}

Java




// Java program to make three non-empty sets
// as per the given condition.
import java.util.*;
 
class GFG
{
 
    static void makeSets(int arr[], int n)
    {
        Vector<Integer> first = new Vector<Integer>();
        Vector<Integer> second = new Vector<Integer>();
        Vector<Integer> third = new Vector<Integer>();
 
        // insert number equal to 0 to third set.
        // numbers greater than 0 to second set.
        // insert numbers less than 0 to first set.
        for (int i = 0; i < n; i++)
        {
            if (arr[i] == 0)
            {
                third.add(arr[i]);
            }
            if (arr[i] > 0)
            {
                second.add(arr[i]);
            }
            if (arr[i] < 0)
            {
                first.add(arr[i]);
            }
        }
 
        if (first.size() == 0 || third.size() == 0)
        {
            System.out.print("Not Possible");
            return;
        }
 
        // if second set is empty.
        if (second.size() == 0)
        {
            for (int i = 0; i < 2; i++)
            {
                second.add(first.lastElement());
                first.remove(first.size() - 1);
            }
        }
 
        // if length of first set is even.
        if (first.size() % 2 == 0)
        {
            third.add(first.lastElement());
            first.remove(first.size() - 1);
        }
 
        // output the first set elements.
        for (int i = 0; i < first.size(); i++)
        {
            System.out.print(first.get(i) + " ");
        }
 
        // output the second set elements.
        System.out.println();
        for (int i = 0; i < second.size(); i++)
        {
            System.out.print(second.get(i) + " ");
        }
 
        // output the third set elements.
        System.out.println();
        for (int i = 0; i < third.size(); i++)
        {
            System.out.print(third.get(i) + " ");
        }
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = {-1, 2, 0};
        int n = arr.length;
        makeSets(arr, n);
    }
}
 
/* This code contributed by PrinciRaj1992 */

Python3




# Python3 program to make three non-empty sets
# as per the given conditions.
def makeSets(arr, n):
    first = []
    second = []
    third = []
     
    # insert number equal to 0 to third set.
    # numbers greater than 0 to second set.
    # insert numbers less than 0 to first set.
    for i in range(n):
        if (arr[i] == 0):
            third.append(arr[i])
        if (arr[i] > 0):
            second.append(arr[i])
        if (arr[i] < 0):
            first.append(arr[i])
     
    if (len(first) == 0 or len(third) == 0):
        print("Not Possible")
        return
     
    # if second set is empty.
    if (len(second)== 0):
        for i in range(2):
            second.append(first[-1])
            first.pop()
             
    # if length of first set is even.
    if (len(first) % 2 == 0):
        third.append(first[-1])
        first.pop()
     
    # output the first set elements.
    for i in range(len(first)):
        print(first[i], end = " ")
     
    # output the second set elements.
    print()
    for i in range(len(second)):
        print(second[i], end = " ")
         
    # output the third set elements.
    print()
    for i in range(len(third)):
        print(third[i], end = " ")
         
# Driver Function
arr = [-1, 2, 0]
n = len(arr)
makeSets(arr, n)
 
# This Code is contributed by SHUBHAMSINGH10

C#




// C# program to make three non-empty sets
// as per the given condition.
using System;
using System.Collections.Generic;
 
class GFG
{
 
    static void makeSets(int []arr, int n)
    {
        List<int> first = new List<int>();
        List<int> second = new List<int>();
        List<int> third = new List<int>();
 
        // insert number equal to 0 to third set.
        // numbers greater than 0 to second set.
        // insert numbers less than 0 to first set.
        for (int i = 0; i < n; i++)
        {
            if (arr[i] == 0)
            {
                third.Add(arr[i]);
            }
            if (arr[i] > 0)
            {
                second.Add(arr[i]);
            }
            if (arr[i] < 0)
            {
                first.Add(arr[i]);
            }
        }
 
        if (first.Count == 0 || third.Count == 0)
        {
            Console.Write("Not Possible");
            return;
        }
 
        // if second set is empty.
        if (second.Count == 0)
        {
            for (int i = 0; i < 2; i++)
            {
                second.Add(first[first.Count-1]);
                first.Remove(first.Count - 1);
            }
        }
 
        // if length of first set is even.
        if (first.Count % 2 == 0)
        {
            third.Add(first[first.Count-1]);
            first.Remove(first.Count - 1);
        }
 
        // output the first set elements.
        for (int i = 0; i < first.Count; i++)
        {
            Console.Write(first[i] + " ");
        }
 
        // output the second set elements.
        Console.WriteLine();
        for (int i = 0; i < second.Count; i++)
        {
            Console.Write(second[i] + " ");
        }
 
        // output the third set elements.
    Console.WriteLine();
        for (int i = 0; i < third.Count; i++)
        {
            Console.Write(third[i] + " ");
        }
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        int []arr = {-1, 2, 0};
        int n = arr.Length;
        makeSets(arr, n);
    }
}
 
// This code has been contributed by 29AjayKumar

Javascript




<script>
 
// JavaScript program to make three non-empty sets
// as per the given conditions.
 
 
function makeSets(arr, n) {
    let first = [], second = [], third = [];
 
    // insert number equal to 0 to third set.
    // numbers greater than 0 to second set.
    // insert numbers less than 0 to first set.
    for (let i = 0; i < n; i++) {
        if (arr[i] == 0)
            third.push(arr[i]);
        if (arr[i] > 0)
            second.push(arr[i]);
        if (arr[i] < 0)
            first.push(arr[i]);
    }
 
    if (first.length == 0 || third.length == 0) {
        document.write("Not Possible");
        return;
    }
 
    // if second set is empty.
    if (second.length == 0) {
        for (let i = 0; i < 2; i++) {
            second.push(first[first.length - 1]);
            first.pop();
        }
    }
 
    // if length of first set is even.
    if (first.length % 2 == 0) {
        third.push(first[first.length - 1]);
        first.pop();
    }
 
    // output the first set elements.
    for (let i = 0; i < first.length; i++)
        document.write(first[i] + " ");
 
    // output the second set elements.
    document.write("<br>");
    for (let i = 0; i < second.length; i++)
        document.write(second[i] + " ");
 
    // output the third set elements.
    document.write("<br>");
    for (let i = 0; i < third.length; i++)
        document.write(third[i] + " ");
}
 
// Driver Function
 
let arr = [-1, 2, 0];
let n = arr.length;
makeSets(arr, n);
 
</script>

Output:  

-1 
2 
0

Time Complexity :- O(n)
 

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